Problem: The base of a solid $S$ is the region bounded by the graphs of the functions $f$ and $g$ shown below. $(c,d)$ $ f$ $ g$ $y$ $x$ Each cross-section perpendicular to the $x$ -axis is an isosceles right triangle with the hypotenuse in the base of solid $S$. Which one of these integrals represents the volume of solid $S$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac14\int_0^c\,[f(x)-g(x)]^2dx$ (Choice B) B $\dfrac12\int_0^d\left[g^{-1}(y)-f^{-1}(y)\right]^2dy$ (Choice C) C $\dfrac12\int_0^c\,[f(x)-g(x)]^2dx$ (Choice D) D $\dfrac14\int_0^d\left[g^{-1}(y)-f^{-1}(y)\right]^2dy$ (Choice E) E None of the above
Explanation: The base of the solid is shaded in blue. Let's add a thin orange rectangle to depict a representative cross-section sitting on the base. The length of the green segment is $b$. $(c,d)$ $ f$ $ g$ $b$ $y$ $x$ Since each cross-section is perpendicular to the $x$ -axis, the independent variable is $x$. Given an $x$ -value, we can calculate the length $b$ from the functions $f$ and $g$ via the equation $b=f(x)-g(x)$. We can see from the graph that $x$ goes from $0$ to $c$. If $A$ denotes the area of each cross-section as a function of $x$, the volume $V$ of solid $S$ is $ V=\int_0^c A(x) \,dx$. To determine the area $A$ as a function of $x$, first express $A$ in terms of $b$. Since the triangular cross-section rests on the rectangle pictured above, the length of the base of the triangle is $b$. Since the triangle is right isosceles with hypotenuse at the base, the height of the triangle is $b/2$. $\dfrac b2$ $b$ The area $A$ of the triangle is $A=\dfrac12\cdot b\cdot\dfrac b2=\dfrac14b^2$. What is $A$ as a function of $x$ ? Let's substitute $b=f(x)-g(x)$ into $A=\dfrac14b^2$ to get $A(x)=\dfrac14[f(x)-g(x)]^2$. Can you express the volume $V$ of solid $S$ as a definite integral? The volume formula gives us the definite integral $\begin{aligned} V&=\int_0^c A(x) \,dx \\\\ &=\int_0^c \dfrac14[f(x)-g(x)]^2 \,dx \\\\ &=\dfrac14\int_0^c [f(x)-g(x)]^2 \,dx \end{aligned}$